4.11 The Inverse and Determinant of a Matrix

The identity matrix I is square, with 1s on the main diagonal (top left to bottom right) and 0s everywhere else. The 2 × 2 identity is [1 0; 0 1].

Multiplying a square matrix by the identity of the same size leaves it unchanged: AI = IA = A.

For A = [a b; c d], the determinant is det(A) = ad − bc. Multiply down the main diagonal and subtract the product of the other diagonal.

If the two columns (or the two rows) of A are vectors in the plane, the absolute value of det(A), when it isn't 0, is the area of the parallelogram they span.

If det(A) = 0, the two vectors lie on the same line through the origin (or one of them is the zero vector), so they don't span any area.

The sign of the determinant tells you the order of the two vectors: positive if the second column is counterclockwise from the first (turning less than half a turn), negative if it's clockwise. The area is the absolute value either way.

The inverse of a square matrix A, written A⁻¹, satisfies AA⁻¹ = A⁻¹A = I. A has an inverse exactly when det(A) ≠ 0.

For a 2 × 2 matrix A = [a b; c d] with ad − bc ≠ 0: A⁻¹ = (1/(ad − bc)) [d −b; −c a]. Swap the diagonal entries, change the signs of the other two, and divide every entry by the determinant.

For larger matrices, use technology. You can always check an inverse by multiplying: the product should be I.

Just as dividing undoes multiplying, multiplying by A⁻¹ undoes multiplying by A. If Ax = b, then x = A⁻¹b. This is how you work backward, as in topics 4.13 and 4.14.

For example, the system 4x + 7y = 5 and 2x + 6y = 0 can be written as A[x; y] = [5; 0] with A = [4 7; 2 6]. Using A⁻¹ = [0.6 −0.7; −0.2 0.4] from the first worked example, [x; y] = A⁻¹[5; 0] = [0.6(5) − 0.7(0); −0.2(5) + 0.4(0)] = [3; −1]. Check: 4(3) + 7(−1) = 5 and 2(3) + 6(−1) = 0.